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Q_sp- ion-product expression; Q_sp value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp value is called K_sp value (solubility-product constant).

${\mathbf{\text{MX}}}_{\mathbf{\text{2}}}\mathbf{→}{\mathbf{\text{M}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{+ 2X}}}^{\mathbf{\text{-}}}$

Solid and liquid state is not included in the K_spequations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{= [M}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{][X}}}^{\mathbf{\text{-}}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S (Molar solubility) is equal to the concentration of one mol of the ion formed by dissolution of some compound.

• Molarity of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$is: $\text{0.155 M}$
• Volume of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ is: $100\mathrm{mL}=0.1\mathrm{L}$
• Volume of Blood is: $\text{104 L}$
• Amount of ${\text{Ca}}^{\text{2 +}}$ion in blood is: $9.7×{10}^{-5}\mathrm{g}/\mathrm{mL}$

First, the dissolution equation for ${\text{CaC}}_{\text{2}}{\text{O}}_4$–

${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s)}⇌{\text{Ca}}^{\text{2 +}}{\text{(aq) +C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Now find the molarity of ${\text{Ca}}^{\text{2 +}}$ions –

$$\begin{gathered} \mathrm{n}\left({\mathrm{Ca}}^{2+}\right)=9.7×{10}^{-5}\mathrm{g}/\mathrm{ml}×104\mathrm{ml}×\frac{1\mathrm{mol}}{40.08\mathrm{g}} \\ \mathrm{n}\left({\mathrm{Ca}}^{2+}\right)=2.5×{10}^{-4}\mathrm{mol} \\ \mathrm{n}\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right)=0.1\mathrm{L}×0.155\mathrm{M} \\ \mathrm{n}\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right)=0.0155\mathrm{mol} \end{gathered}$$

Now subtract ${\text{n(Ca}}^{\text{2 +}})$from ${\text{n(C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2 -}})$–

$0.0155\mathrm{mol}-0.00025\mathrm{mol}=0.01525\mathrm{mol}$

Now calculate the total volume of solution –

$0.1\mathrm{L}+0.104\mathrm{L}=0.204\mathrm{L}$

Now, calculate concentration of ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2 -}}$ions –

$$\begin{gathered} \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=\frac{0.01525\mathrm{mol}}{0.204\mathrm{L}} \\ =7.4×{10}^{-2}\mathrm{M} \end{gathered}$$

The K_sp value is –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right] \\ \left[{\mathrm{Ca}}^{2+}\right]=\mathrm{S} \\ \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=7.4×{10}^{-2} \\ {\mathrm{K}}_{\mathrm{sp}}=2.3×{10}^{-9} \end{gathered}$$

Number of ${\text{Ca}}^{\text{2 +}}$ions left are –

$$\begin{gathered} 2.3×{10}^{-9}=\mathrm{S}×\left(7.4×{10}^{-2}\right) \\ \left[{\mathrm{Ca}}^{2+}\right]=3.108×{10}^{-8}\mathrm{M} \end{gathered}$$

Therefore, the value for number of moles is obtained as $3.108×{10}^{-8}\mathrm{M}$.